The AM of the series 1, 2, 4, 8, 16, . . ., 2n is

A
$\frac{2^{n} - 1}{n}$
B
$\frac{2^{n + 1} - 1}{n + 1}$
C
$\frac{2^{n} + 1}{n}$
D
$\frac{2^{n} - 1}{n + 1}$

Solution:

The required AM is
$\bar{X} = \frac{1 + 2 + 2^{2} + 2^{3} + \dots + 2^{n}}{n + 1} = \frac{(2^{n + 1} - 1)}{(n + 1) (2 - 1)} = \frac{2^{n + 1} - 1}{n + 1}$

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